Function f(x) = frac{|x - 1|}{x^2} is monoton | vedclass

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(D) The function is defined as $f(x) = \begin{cases} \frac{-(x-1)}{x^2} & x < 1 \ \frac{x-1}{x^2} & x \geq 1 \end{cases}$.Calculating the derivative

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$(0, 1) \cup (2, \infty)$

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(D) The function is defined as $f(x) = \begin{cases} \frac{-(x-1)}{x^2} & x Calculating the derivative $f'(x)$:For $x For $x > 1$,$f(x) = \frac{x-1}{x^2} = x^{-1} - x^{-2}$,so $f'(x) = -x^{-2} + 2x^{-3} = \frac{-(x-2)}{x^3}$.For $f(x)$ to be monotonically decreasing,we require $f'(x) Case $1$: $x Case $2$: $x > 1$. We need $\frac{-(x-2)}{x^3}  0$. This holds when $x \in (-\infty, 0) \cup (2, \infty)$. Combining with $x > 1$,we get $x \in (2, \infty)$.Therefore,the function is monotonically decreasing in $(0, 1) \cup (2, \infty)$.

Function $f(x) = \frac{|x - 1|}{x^2}$ is monotonically decreasing in

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